The Weirstrass Test for absolute uniform convergence of improper integrals states that if there is a function $M(x)$ such that $|f(x,y)| \leq M(x)$ for all $y$,$x$ in the domain of consideration, then if $\int_a^b M(x)dx < \infty$ we have that $\int_a^b f(x,y)dx$ converges absolutely/uniformly.

So if $f(x,y)\geq 0$ for all $x,y$ in the domain of interest then I can just choose $M(x) = f(x,y)$. Then if $\int_a^bf dx$ converges then I'm done right?

I feel that I'm missing something here since I haven't seen this stated anywhere.

  • $\begingroup$ Where is $y$ coming from? This is very hard to follow. But if you have $f\ge 0$, then absolute convergence is the same as convergence. $\endgroup$ Commented Feb 21, 2018 at 23:12
  • $\begingroup$ Sorry. $f$ is a multivariate function, and $y$ is an argument. I know that convergence in this case implies absolute convergence, but it seems that it also implies uniform convergence. $\endgroup$
    – Ten Times
    Commented Feb 21, 2018 at 23:15
  • $\begingroup$ Please be very precise about what improper integral you're considering. Are you integrating over a region in the plane? What makes things improper? $\endgroup$ Commented Feb 21, 2018 at 23:17

1 Answer 1


Choosing $M(x) = f(x,y)$ doesn't make sense, because $f(x,y)$ depends on $x$ and $y$ but $M(x)$ may only depend on $x$.

You may certainly define $M(x) = \sup_y f(x,y)$, which (when it's finite, which doesn't happen all the time) does supply an upper bound suitable for the Weierstrass test.


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